Question

Let the function $f:R\to R$ be defined by $f\left(x\right)=cosx,\forall x\in R$. Show that $f$. is neither one-one nor onto.

Answer 1

For a function f: X → Y

f is one-one (or injective), if distinct
elements of X have distinct images in Y.

$i.e.,$ for every $x_1,x_2\in X,$
$x_1\ne x_2\Rightarrow f\left(x_1\right)\ne f\left(x_2\right)$

$f\left(x_1\right)=f\left(x_2\right)\Rightarrow x_1=x_2$


Here, $f\left(x\right)=cosx\forall x\in R$

$ƒ(-\frac{\pi }{2})=cos(-\frac{\pi }{2})=0$

Also, $f\left(\frac{\pi }{2}\right)$ =cos $\left(\frac{\pi }{2}\right)$

$\Rightarrow (-\frac{\pi }{2})$ = $f(-\frac{\pi }{2})$

But $\frac{\pi }{2}\ne \frac{\pi }{2}\therefore f\left(x\right)$ is not one-one

Solution:

Now, checking if the function is onto.

For that finding range of the function.

$f\left(x\right)=cosx\forall x\in Rwheref:R\to R$

Range of $cosx=[-1,1]$

And Co-domain = $R$

$\text{Since range}\ne \text{co-domain}$

So, the given function is not onto.

$\therefore$ Given function is neither one-one nor onto.

Hence proved.